Estimating $\sin(-0.1)$ using a Maclaurin polynomial, what is the least degree of the polynomial that assures an error smaller than $0.001$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $3$ (Choice C) C $4$ (Choice D) D $5$
Solution: We will use the Lagrange error bound. Let's assume the polynomial's degree is $n$. The absolute value of the $(n+1)^{\text{th}}$ derivative of $\sin(x)$ is bounded by $1$. [Why?] The Lagrange bound for the error assures that: $\begin{aligned}|R_n(-0.1)| &\leq \left| \dfrac{1}{(n+1)!}(-0.1)^{n+1} \right| \\\\ &= \dfrac{0.1^{n+1}}{(n+1)!} \end{aligned}$ Solving $\dfrac{0.1^{n+1}}{(n+1)!}<0.001$ using trial and error, we find that $n\geq2$. In conclusion, the least degree of the polynomial that assures our error bound is $2$.